1
$\begingroup$

Let us say I am given a starting (latitude, longitude)=(lat,lon) coordinate in degrees. The objective is to compute the new latitude, lat' when moving d miles north along a fixed longitude, lon.

More specifically, I want a formula for lat' given an initial (lat, lon) and a distance in miles (or kilometers), d on how long one will move North across the fix starting longitude lon.

Ideally, we'd take into account the fact that the earth is an ellipsoid. A rough approximation will also work.

Also for simplification let us assume that one is not too close to the poles (100+ miles away from the poles) and that the distance d is less than 20 miles north - this will avoid the need to handle edge cases.

Thanks!

$\endgroup$

1 Answer 1

1
$\begingroup$

If your distances are fairly short like this, we can use the formula for the meridian arc $M(\phi_o)$ to determine the local meridian radius of curvature at the starting latitude $\phi_o$, and then use it as a basis for the calculation of the latitude of destination $\phi_d$.

$$M(\phi_o) = \frac{(ab)^2}{\left(\left(a \text{ cos } \phi_o \right)+ \left(b \text{ sin } \phi_o\right)^2\right)^\frac{3}{2}}$$ $$\phi_d = \phi_o + d/M(\phi_o)$$

Where $a$ is the semi major-axis and $b$ is the semi minor-axis of the ellipsoid. All angles are in radians.

Finally, calculate the midpoint latitude (average of $\phi_o$ and $\phi_d$) and recalculate the radius of curvature with this new value, then use this result in the second formula instead of the old $M(\phi_o)$ to get the final answer.

With this method, for a distance of 20 miles, the resulting destination latitude will always be within 1 millimeter of the real answer.

Of course, an much simpler but less accurate method would be to assume that the Earth is spherical, and use a constant Earth radius: $\phi_d = \phi_o + d/r$ This can generate a maximum error of about 0.5% of the distance.

This Wikipedia article explains the problem in more detail and how to solve it more accurately with integrals and iterative solutions.

$\endgroup$
1
  • $\begingroup$ this is a very simple, accurate and great approximation, thanks for the answer! $\endgroup$
    – Dnaiel
    Commented Mar 15, 2018 at 13:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .